Showing A Set is Vector Space

Question

• Check that the set $\mathbb{R}^\omega$ of all infinite-tuples of real numbers

$$x=(x_1,x_2,...)$$

is a vector space under component-wise addition and scalar multiplication.

My Attempt. Let $x,y\in\mathbb{R}^\omega$. We need to show:

1) $x+y=(x_1,x_2,...)+(y_1,y_2,...)=(x_1+y_1,x_2+y_2,...)\in\mathbb{R}^\omega$

2)$cx=c(x_1,x_2,...)=(cx_1,cx_2,...)\in\mathbb{R}^\omega$.

How do we know $1)$ and $2)$ is true? Can you explain, can you show $1),2)?$

Answer 1

At first we have to explain why $(x_1 + y_1, x_2 + y_2, \dots)$ and $(cx_1, cx_2, \dots)$ are elements of $\mathbb R^\omega$. Note that this is true because the sum of two real numbers is again a real number and the product of two real numbers is again a real number.

So what we now have to do is to verify the eight vector space axioms (see e. g. the wiki page). I will give you here the proof of two of them and let you the proof of the remaining ones as an exercise.

Commutativity of addition: Let $x = (x_1, x_2, \dots), y = (y_1, y_2, \dots)$ as above. Then

$$x + y = (x_1 + y_1, x_2 + y_2, \dots)\ {\color{red} = }\ (y_1 + x_1, y_2 + x_2, \dots) = y + x.$$

Here, the crucial point is to explain why the red marked equality holds. This is the case because of the commutativity of the addition in $\mathbb R$: We have $a + b = b + a $ for all $a, b \in \mathbb R$. (Remark: You will need the properties of $\mathbb R$ for showing that $\mathbb R^\omega$ is a vector space, everything follows from this.)

Existence of a zero vector: We have to show that there is a $0_{\mathbb R^\omega}$ such that $x + 0_{\mathbb R^\omega} = x = 0_{\mathbb R^\omega} + x$ for all $x \in \mathbb R^\omega$. Note that I write $0_{\mathbb R^\omega} $ more because of pedagogical reasons, this distinguishes the zero vector from $0 \in \mathbb R$. So take

$$0_{\mathbb R^\omega} = (0, 0, \dots) \in \mathbb R^\omega.$$

Then

\begin{align} x + 0_{\mathbb R^\omega} &= (x_1 + 0, x_2 + 0, \dots ) = (x_1, x_2, \dots) = x , \\ 0_{\mathbb R^\omega} + x &= (0 + x_1, 0 + x_2, \dots ) = (x_1, x_2, \dots) = x. \end{align}

Here we use again a fact of the real numbers: The zero element is there $0$, so we have $x_1 + 0 = 0 + x_1 = x_1$ for every real $x_1$. We have established the existence of a zero vector, you now have to prove the remaining six axioms.

Answer 2

Let $x,y\in\mathbb{R}^\omega$. Since we know 1) $x+y=(x_1,x_2,...)+(y_1,y_2,...)=(x_1+y_1,x_2+y_2,...)\in\mathbb{R}^\omega$

2)$cx=c(x_1,x_2,...)=(cx_1,cx_2,...)\in\mathbb{R}^\omega$,

then $\mathbb{R}^\omega$ is a vector space.